Find the 2nd largest element in an array with minimum number of comparisons

Question

For an array of size N, what is the number of comparisons required?

Answer 1

    int[] int_array = {4, 6, 2, 9, 1, 7, 4, 2, 9, 0, 3, 6, 1, 6, 8};
    int largst=int_array[0];
    int second=int_array[0];
    for (int i=0; i<int_array.length; i++){        
        if(int_array[i]>largst) { 
            second=largst;
            largst=int_array[i];
        }  
        else if(int_array[i]>second  &&  int_array[i]<largst) { 
            second=int_array[i];
        } 
    }

Answer 2

Use counting sort and then find the second largest element, starting from index 0 towards the end. There should be at least 1 comparison, at most n-1 (when there's only one element!).

Answer 3

package com.array.orderstatistics;

import java.util.Arrays;
import java.util.Collections;

public class SecondLargestElement {

    /**
     *  Total Time Complexity will be n log n + O(1)
     * @param str
     */
    public static void main(String str[]) {
        Integer[] integerArr = new Integer[] { 5, 1, 2, 6, 4 };



        // Step1 : Time Complexity will be n log(n)
        Arrays.sort(integerArr, Collections.reverseOrder());

        // Step2 : Array.get Second largestElement
        int secondLargestElement = integerArr[1];

        System.out.println(secondLargestElement);
    }
}

Answer 4

Suppose provided array is inPutArray = [1,2,5,8,7,3] expected O/P -> 7 (second largest)

 take temp array 
      temp = [0,0], int dummmy=0;
    for (no in inPutArray) {
    if(temp[1]<no)
     temp[1] = no
     if(temp[0]<temp[1]){
    dummmy = temp[0]
    temp[0] = temp[1]
    temp[1] = temp
      }
    }

    print("Second largest no is %d",temp[1])

Answer 5

I have gone through all the posts above but I am convinced that the implementation of the Tournament algorithm is the best approach. Let us consider the following algorithm posted by @Gumbo

largest := numbers[0];
secondLargest := null
for i=1 to numbers.length-1 do
    number := numbers[i];
    if number > largest then
        secondLargest := largest;
        largest := number;
    else
        if number > secondLargest then
            secondLargest := number;
        end;
    end;
end;

It is very good in case we are going to find the second largest number in an array. It has (2n-1) number of comparisons. But what if you want to calculate the third largest number or some kth largest number. The above algorithm doesn't work. You got to another procedure.

So, I believe tournament algorithm approach is the best and here is the link for that.

Answer 6

class solution:
def SecondLargestNumber (self,l):
    Largest = 0
    secondLargest = 0
    for i in l:
        if i> Largest:
            secondLargest = Largest
        Largest = max(Largest, i)
    return Largest, secondLargest